已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)(1)求最小正周期(2)求函数单调增区间(3)若x属于【0,pai/2】时,F(x)的最小值是-2,求a
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![已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)(1)求最小正周期(2)求函数单调增区间(3)若x属于【0,pai/2】时,F(x)的最小值是-2,求a](/uploads/image/z/3793891-67-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%CF%80%2F6%29%2Bsin%282x%E2%80%94%CF%80%2F6%29%E2%80%94cos2x%2Ba%28a%E4%B8%BA%E5%AE%9E%E6%95%B0%2C%E5%B1%9E%E4%BA%8ER%29%EF%BC%881%EF%BC%89%E6%B1%82%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0%E5%8D%95%E8%B0%83%E5%A2%9E%E5%8C%BA%E9%97%B4%EF%BC%883%EF%BC%89%E8%8B%A5x%E5%B1%9E%E4%BA%8E%E3%80%900%2Cpai%2F2%E3%80%91%E6%97%B6%2CF%EF%BC%88x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%98%AF-2%2C%E6%B1%82a)
已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)(1)求最小正周期(2)求函数单调增区间(3)若x属于【0,pai/2】时,F(x)的最小值是-2,求a
已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)
(1)求最小正周期
(2)求函数单调增区间
(3)若x属于【0,pai/2】时,F(x)的最小值是-2,求a
已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)(1)求最小正周期(2)求函数单调增区间(3)若x属于【0,pai/2】时,F(x)的最小值是-2,求a
第一个问题:
f(x)=sin2xcos(π/6)+cos2xsin(π/6)+sin2xcos(π/6)-cos2xsin(π/6)-cos2x+a
=2sin2xcos(π/6)-cos2x+a=2[sin2xcos(π/6)-cos2xsin(π/6)]+a
=2sin(2x-π/6)+a.
∴函数f(x)的最小正周期为2π/2=π.
第二个问题:
∵f(x)=2sin(2x-π/6)+a.∴当 2kπ-π/2≦2π-π/6≦2kπ+π/2 时,f(x)单调递增.
由2kπ-π/2≦2x-π/6≦2kπ+π/2,得:2kπ-3π/6+π/6≦2x≦2kπ+3π/6+π/6,
∴2kπ-2π/6≦2x≦2kπ+4π/6,∴kπ-π/6≦x≦kπ+π/3.
即函数f(x)的单调增区间是[kπ-π/6,kπ+π/3],其中k为整数.
第三个问题:
∵0≦x≦π/2 ,∴0≦2x≦π,∴-π/6≦2x-π/6≦π-π/6,
∴f(x)的最小值为2sin(-π/6)+a=-1+a=-2,∴a=-1.
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